Ex 5.3, 4 - How many terms of AP 9, 17, 25 … must be - Finding number of terms given s

Ex 5.3 part 1.jpg

  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
Ask Download

Transcript

Ex 5.3 ,4 How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636? AP is 9,17,25, ….. We know that Sum = 𝑛/2 (2a + (n – 1) d) Here, a = 9 d = 17 – 9 = 8 & Sum = Sn = 636 We need to find n Putting values in equation Sum = 𝑛/2 (2a + (n – 1) d) 636 = 𝑛/(2 ) (2 Γ— 9 + (n – 1) Γ— 8) 636 Γ— 2 = n (18 + (n – 1) Γ— 8) 1272 = n (18 + 8n – 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 – 1272 10n + 8n2 – 1252 = 0 2(5n + 4n2 – 636)= 0 5n + 4n2 – 636 = 0 4n2 + 5n – 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = – 636 We know that, D = b2 – 4ac D = (5)2 – 4 Γ— 4 Γ— (–636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (βˆ’ 𝑏 Β± √𝐷)/2π‘Ž Putting values n = (βˆ’ 5 Β± √10201)/(2 Γ— 4) n = (βˆ’ 5 Β± 101)/8 Solving So, n = 12 & n = (βˆ’53)/4 Since n cannot be negative, n = 12

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.